#TotalProbability

[sethiswiftlearn]

Class 12 Physics | Magnetic effects of current | JEE Mains, NEET, CUET Preparation

Probability is one of the most important topics in Class 12 Mathematics, especially for students preparing for competitive exams like JEE Mains, CUET, and other entrance tests. Understanding Bayes’ Theorem and Theorems on Total Probability can significantly boost your problem-solving skills and help you tackle probability-based questions with confidence.

In this detailed session, we break down the concepts of Bayes' Theorem and the Total Probability Theorem with easy-to-understand explanations and solved examples. This video is perfect for students who want to strengthen their conceptual clarity and apply probability in real-world scenarios.

🔹 What You Will Learn in This Video: ✔️ Understanding Conditional Probability ✔️ Theorem on Total Probability – Definition and Application ✔️ Bayes' Theorem – Concept, Derivation, and Formula ✔️ Step-by-step Problem-Solving Approach ✔️ JEE Mains & CUET Level Questions with Solutions ✔️ Real-life applications of Bayes’ Theorem

Bayes’ Theorem is an essential topic in probability that helps in making data-driven decisions by updating prior probabilities based on new evidence. Many real-world applications, such as medical testing, spam filtering, and artificial intelligence, rely on this theorem. That’s why mastering this concept is crucial for students aiming to excel in competitive exams and those interested in fields like Data Science and Machine Learning.

🔥 Why Should You Watch This Video?

✅ Conceptual Clarity: We simplify complex topics with visual explanations and real-world applications. ✅ Exam-Focused Approach: Covers frequently asked questions in JEE Mains, CUET, and board exams. ✅ Step-by-Step Solutions: Learn problem-solving techniques that improve accuracy and speed. ✅ Expert Guidance: Learn from experienced educators who provide shortcuts and tricks to tackle tricky probability questions.

🚀 Start Learning Now! Click the link below to watch the full session: 🎥 Watch here: https://youtu.be/muCa3PTx7eA?si=KV5kKDXNbpnQXLut

📢 Who Should Watch This?

📌 JEE Mains & Advanced Aspirants 📌 CUET & Other Entrance Exam Students 📌 Class 12 CBSE & State Board Students 📌 Anyone Looking to Master Probability Concepts

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[statprofzhu]

A simple formula to quantify bias in opinion polls

\(\qquad\) The election of Donald J. Trump as the 45th President of the United States in 2016 shocked many people, especially since most opinion polls prior to the election consistently predicted his defeat. It turned out that many of his supporters didn't feel comfortable expressing their true views. One does not need a degree in statistics to understand that, if his critics are much more likely than his supporters to be included in opinion polls, then the estimates coming out of those polls will be biased downwards. This qualitative conclusion is a no-brainer, but is there an easy way to quantify the amount of the bias?

\(\qquad\) Here is a simple formula which one can use to perform some back-of-the-envelope calculations. Let \(\theta\) denote the true proportion of those who support a certain candidate. If an opinion poll is \(k\) times as likely to include those who do not support him as those who do, then on average it will estimate \(\theta\) to be \[ \bar{\theta}_{\text{poll}}=\theta/[k-(k-1)\theta]. \] For transparency, the mathematical derivation of this simple formula is provided below in the Appendix, so that it may be checked by those who have the desire to do so. For most other readers, a simple numeric example will suffice to illustrate how it can be applied.

\(\qquad\) Suppose, hypothetically, that \(54\%\) of Americans actually supported Trump in 2016 (i.e., \(\theta=0.54\)), but those who did not support him were, for whatever reason, twice as likely to end up in opinion polls as those who did (i.e., \(k=2\)). Then, on average the polls would have estimated that only \((0.54)/[2-(2-1)(0.54)] \approx 0.37\)---or about \(37\%\)---of Americans appeared to support him, giving a downward bias of a whopping \(17\) percentage points!

\(\qquad\) Of course, with a good guess of what the value of \(k\) is, one can also use this formula to infer the true proportion \(\theta\) from the poll estimate by \(k\bar{\theta}_{\text{poll}}/[(k-1)\bar{\theta}_{\text{poll}}+1]\), e.g., \(2(0.37)/[(2-1)(0.37)+1] \approx 0.54\). Moreover, after the election is over and the true proportion \(\theta\) is revealed, one can even use this formula to obtain a post-hoc estimate of \(k\) as \([\theta/(1-\theta)]/[\bar{\theta}_{\text{poll}}/(1-\bar{\theta}_{\text{poll}})]\), e.g., \([(0.54)/(1-0.54)]/[(0.37)/(1-0.37)] \approx 2\). An ideal value is clearly \(k=1\). What it means sociologically to have a significant deviation from this ideal is anyone's speculation.

(by Professor Z, November 2020)

Appendix \( \def\S{\mathcal{S}} \def\E{\mathbb{E}} \def\pr{\mathbb{P}} \)

\(\qquad\) Suppose there is a finite population of fixed values, \(\{x_{1}, ..., x_{N}\\}\), and we simply want to know what their average, \[ \theta=\frac{1}{N} \sum_{i=1}^{N} x_{i}, \] is equal to. We cannot afford to query every value because \(N\) is too large, so we take a sample of size \(n \ll N\), e.g., \(\{x_{3}, x_{15}, ..., x_{523}\}\), in which case we shall denote the corresponding index set---here, \(\{3, 15, ..., 523\}\)---by \(\S\). We simply estimate the unknown quantity \(\theta\) by its "sample analogue", that is, by \[ \widehat{\theta} = \frac{1}{n} \sum_{i\in\S} x_i. \] Clearly, there is inherent randomness in the quantity \(\widehat{\theta}\) depending on which values end up in our sample \(\S\), so we are asking instead what it will turn out to be on average. That is, we want the value of \(\bar{\theta}_{\text{poll}} \equiv \E(\widehat{\theta})\).

\(\qquad\) Now consider the case in which the population can be partitioned into two groups, \[ \{ \underbrace{x_{1}, ..., x_{A},}_{\text{Group 1}} \quad \underbrace{x_{A+1}, ..., x_{N}}_{\text{Group 2}} \}, \] and the probability that each value will end up in our sample, \(\pr(i\in\S)\), differs depending on whether \(i\) belongs to the first group (\(i\leq A\)) or the second one (\(i > A\)). We can rewrite \[ \theta=\frac{1}{N}\left[\sum_{i=1}^A x_i + \sum_{i=A+1}^N x_i\right] \tag{1} \label{eq:theta-twogroup} \] to reflect the group structure explicitly and, for each \(i\), define two (conditional) indicators, \(S^{-}_i\) and \(S^{+}_i\), respectively for \(i \leq A\) and \(i > A\), such that \[ \pr(S^{-}_i=1) \equiv \pr(i\in\S|i \leq A)\quad\text{and}\quad \pr(S^{+}_i=1) \equiv \pr(i\in\S|i > A). \] This allows us to express our sample estimator \(\widehat{\theta}\) as \[ \widehat{\theta}=\frac{1}{n} \left[\sum_{i=1}^A x_i S^{-}_i + \sum_{i=A+1}^N x_i S^{+}_i\right]. \tag{2} \label{eq:thetahat-twogroup} \] By the law of total probability, we must have \[ \underbrace{\pr(i\in\S)}_{n/N} = \underbrace{\pr(i\in\S|i\leq A)}_{p} \underbrace{\pr(i\leq A)}_{A/N} + \underbrace{\pr(i\in\S|i > A)}_{q} \underbrace{\pr(i > A)}_{(N-A)/N}. \tag{3} \label{eq:poll-totalprob} \] One can easily verify from Eq. \(\eqref{eq:poll-totalprob}\) that, if \(q=p\), this simply implies \(p=q=n/N\). So, \[ \E(\widehat{\theta}) =\frac{1}{n} \left[\sum_{i=1}^A x_i\underbrace{\E(S^{-}_i)}_{p=n/N} + \sum_{i=A+1}^N x_i\underbrace{\E(S^{+}_i)}_{q=n/N}\right] \ =\frac{1}{N}\left[\sum_{i=1}^A x_i + \sum_{i=A+1}^N x_i\right] \ =\theta, \] which means \(\hat{\theta}\) is unbiased.

\(\qquad\) We are interested in the following scenario: \[ q=kp \quad\text{for some fixed \(k>1\)} \quad\text{and}\quad x_i = \begin{cases} 1, & i \leq A; \newline 0, & i > A. \end{cases} \] Eq. \(\eqref{eq:theta-twogroup}\) now shows the quantity we are interested in, \(\theta=A/N\), is exactly the proportion of the population belonging to the first group; and Eq. \(\eqref{eq:thetahat-twogroup}\) now implies that \[ \E(\widehat{\theta}) =\frac{1}{n} \left[\sum_{i=1}^A x_i\underbrace{\E(S^{-}_i)}_{p} + \sum_{i=A+1}^N x_i\underbrace{\E(S^{+}_i)}_{q}\right] =\frac{Ap}{n}. \tag{4} \label{eq:Ethetahat-twogroup} \] But if \(q=kp\), we see that \[ p=\frac{n}{kN-(k-1)A} \] by Eq. \(\eqref{eq:poll-totalprob}\), and plugging this into Eq. \(\eqref{eq:Ethetahat-twogroup}\) yields \[ \E(\hat{\theta})=\frac{\theta}{k-(k-1)\theta}, \] our main conclusion.